(solved)Question 2.23 of NCERT Class XI Chemistry Chapter 2

(i) Write the electronic configurations of the following ions: (a) H- (b) Na+ (c) O2- (d) F- (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? (iii) Which atoms are indicated by the following configurations ? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

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Parveen,

Question 2.23
NCERT Class XI Chemistry

(i) Write the electronic configurations of the following ions: (a) H- (b) Na+ (c) O2- (d) F- (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? (iii) Which atoms are indicated by the following configurations ? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

Solution in Detail
(video solution below this)

$\displaystyle \underline{\underline{\text{Part (i)}}}$

calculate electrons first, then configure

$\displaystyle \text{(a) }Z_H = 1\therefore n_e \text{ in H}^-$ = 1 + 1 = 2

$\displaystyle \implies 1s^2 \:\underline{Ans}$

$\displaystyle \text{(b) }Z_{Na} = 11\therefore n_e \text{ in Na}^+$ = 11 - 1 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \text{(c) }Z_{O} = 8\therefore n_e \text{ in O}^{2-}$ = 8 + 2 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \text{(d) }Z_{F} = 9\therefore n_e \text{ in F}^{-}$ = 9 + 1 = 10

$\displaystyle \implies 1s^22s^22p^6\:\underline{Ans}$

$\displaystyle \underline{\underline{\text{Part (ii)}}}$

Complete the previous shells in each case

$\displaystyle \text{(a) }3s^1 \equiv 1s^22s^22p^63s^1 \implies Z = 11$

$\displaystyle \text{(b) }2p^3 \equiv 1s^22s^22p^3 \implies Z = 7$

$\displaystyle \text{(c) }3p^5 \equiv 1s^22s^22p^63s^23p^5 \implies Z = 17$

$\displaystyle \underline{\underline{\text{Part (iii)}}}$

Calculate Z in each case

$\displaystyle \text{(a) [He]}2s^1 $

$\displaystyle \therefore Z = Z_{He} + 1 = 2 + 1 = 3 \text{ Li }\underline{Ans}$

$\displaystyle \text{(b) [Ne]}3s^23p^3 $

$\displaystyle \therefore Z = Z_{Ne} + 5 = 10 + 5 = 15 \text{ P }\underline{Ans}$

$\displaystyle \text{(c) [Ar]}4s^23d^1 $

$\displaystyle \therefore Z = Z_{Ar} + 3 = 18 + 3 = 21 \text{ Sc }\underline{Ans}$

Video Explanation

Please watch this youtube video for a quick explanation of the solution:

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